What is the pH of a 2.6*10^-9 M H^+ solution?

Jul 28, 2016

The pH is 6.99.

Explanation:

In pure water, ["H"^+] = ["OH"^"-"] = 1.00 × 10^"-7" color(white)(l)"mol/L".

Imagine that we add the 2.6 × 10^"-9" color(white)(l)"mol/L H"^+, but the equilibrium hasn't had time to re-establish itself.

We have just set up a new initial condition.

Let's put this into an ICE table.

$\textcolor{w h i t e}{m m m m m} \text{H"_2"O" ⇌ color(white)(mmmmmll)"H"^+color(white)(mmmm) +color(white)(mll) "OH"^"-}$
$\text{I/mol·L"^"-1": color(white)(mmmm)1.00×10^"-7" + 2.6×10^"-9" color(white)(ml)1.00 × 10^"-7}$
$\text{C/mol·L"^"-} : \textcolor{w h i t e}{m m m m m m m m m} - x \textcolor{w h i t e}{m m m m m m m l} - x$
$\text{E/mol·L"^"-1":color(white)(mmmmm) 1.026 × 10^"-7" -xcolor(white)(mml) 1.00 × 10^"-7} - x$

K_w = ["H"^+]["OH"^"-"] = 1.00 × 10^"-14"

(1.026 × 10^"-7" -x)(1.00 × 10^"-7" - x) = 1.00 × 10^"-14"

x^2 - 2.026 × 10^"-7"x + 1.026 × 10^"-14" = 1.00 × 10^"-14"

x^2 - 2.026 × 10^"-7"x + 0.026 × 10^"-14" = 0

x = 1.30 × 10^"-9"

["H"^+] = 1.026 × 10^"-7" -x = 1.026 × 10^"-7" - 1.03 × 10^"-9" = 1.016 × 10^"-7"

"pH" = -log["H"^+] = -log(1.016 × 10^"-7") = 6.99